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kvasek [131]
3 years ago
6

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

n does 288 J of work on the surroundings. What is the final volume of the cylinder? Tro, Nivaldo J.. Chemistry (p. 299). Pearson Education. Kindle Edition.
Physics
1 answer:
Step2247 [10]3 years ago
7 0

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

where

p is the pressure of the gas

V_i is the initial volume

V_f is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

V_i = 0.250 L

And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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DaniilM [7]

\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}

  • The angle between the two vectors is 90° .

\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}

  1. The dot product of two vectors .
  2. The cross product of two vectors .

\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}

⚡ Let \rm{\vec{a}} and \rm{\vec{b}} are the two vectors .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}

  • cos 90° = <u>0</u>

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}

\rm{\red{\therefore}} [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}

  • sin 90° = <u>1</u>

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}

\rm{\red{\therefore}} [2] The cross product of two vectors is “ <u>ab</u> ” .

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3 years ago
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A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls
babunello [35]

Answer:

They are 7.4m apart.

Explanation:

Here we have a parabolic motion problem. we need the time taken to land so:

Y=Yo+Vo*t+\frac{1}{2}*a*t^2

considerating only the movement on Y axis:

0=4.6-(9.81)*t^2\\t=0.68s

Because we have a contant velocity motion on X axis:

xs=vs*t\\xs=17m/s*(0.68s)\\xs=11.6m

and

xg=vg*t\\xg=28m/s*(0.68s)\\xg=19m

the distance between them is given by:

d=|xg-xs|\\d=7.4m

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If a sound wave is produced with a wavelength of 1.04m what is the waves frequency
dmitriy555 [2]
You should just ask the wave 
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The adult house fly lives for only about 1 month, or 8 x 10^-2 y. The oldest recorded age of a tortoise was 1.8800 x 10^2 y. Wha
melisa1 [442]

187.92 years

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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

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3 years ago
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